Решебник по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 6

\[\boxed{\mathbf{6.}}\]

\[\textbf{а)}\ \sqrt[3]{x} = x\]

\[x = x^{3}\]

\[x^{3} - x = 0\]

\[x\left( x^{2} - 1 \right) = 0\]

\[x = 0;\]

\[x^{2} - 1 = 0\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[Ответ:\ x = - 1;0;1.\]

\[\textbf{б)}\ 3\sqrt[3]{x} = x - 2\]

\[27x = (x - 2)^{3}\]

\[27x = x^{3} - 6x^{2} + 12x - 8\]

\[x^{3} - 6x^{2} - 15x - 8 = 0\]

\[x^{3} - 6x^{2} - 7x - 8x - 8 = 0\]

\[x\left( x^{2} - 6x - 7 \right) - 8(x + 1) = 0\]

\[x^{2} - 6x - 7 = (x + 1)(x - 7)\]

\[D_{1} = 9 + 7 = 16\]

\[x_{1} = 3 + 4 = 7;\]

\[x_{2} = 3 - 4 = - 1.\]

\[x(x + 1)(x - 7) - 8(x + 1) = 0\]

\[(x + 1)\left( x^{2} - 7x - 8 \right) = 0\]

\[x + 1 = 0\]

\[x = - 1.\]

\[x^{2} - 7x - 8 = 0\]

\[x_{1} + x_{2} = 7;\ \ x_{1} \cdot x_{2} = - 8\]

\[x_{1} = 8;\ \ x_{2} = - 1.\]

\[Ответ:x = - 1;x = 8.\]

\[\textbf{в)}\ 4\sqrt[3]{x + 2} = x + 2\]

\[64(x + 2) = (x + 2)^{3}\]

\[64(x + 2) - (x + 2)^{3} = 0\]

\[(x + 2)\left( 64 - x^{2} - 4x - 4 \right) = 0\]

\[(x + 2)\left( - x^{2} - 4x + 60 \right) = 0\]

\[x + 2 = 0\]

\[x = - 2.\]

\[- x^{2} - 4x + 60 = 0\]

\[x^{2} + 4x - 60 = 0\]

\[D_{1} = 4 + 60 = 64\]

\[x_{1} = - 2 + 8 = 6;\]

\[x_{2} = - 2 - 8 = - 10.\]

\[Ответ:x = - 10;\ - 2;6.\]

\[\textbf{г)}\ 3\sqrt[3]{x - 2} = x\]

\[27(x - 2) = x^{3}\]

\[27x - 54 - x^{3} = 0\]

\[x^{3} - 27x + 54 + 27 - 27 = 0\]

\[x^{3} - 27x + 81 - 27 = 0\]

\[\left( x^{3} - 27 \right) - (27x - 81) = 0\]

\[(x - 3)\left( x^{2} + 3x + 9 \right) -\]

\[- 27(x - 3) = 0\]

\[(x - 3)\left( x^{2} + 3x + 9 - 27 \right) = 0\]

\[(x - 3)\left( x^{2} + 3x - 18 \right) = 0\]

\[x - 3 = 0\]

\[x = 3.\]

\[x^{2} + 3x - 18 = 0\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 18\]

\[x_{1} = - 6;\ \ x_{2} = 3.\]

\[Ответ:x = - 6;x = 3.\]